A bullet leaves a rifle with a muzzle velocity of 521 m/s. while accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration)

[tex]velocity=\frac{distance}{time}\\\\
velocity*time=distance\\\\
time=\frac{distance}{velocity}\\\\
time=\frac{0,840m}{521\frac{m}{s}}=0,0016s\\\\
acceleration=\frac{velocity}{time}=\frac{521}{0,0016}=325625\frac{m}{s^2}[/tex]

Answer:a = 1.62*105 m /s2Explanation:25.28 m2/s2 = vi2vi = 5.03 m/sTo find hang time, find the time to the peak and then double it.vf = vi + a*t0 m/s = 5.03 m/s + (-9.8 m/s2)*tup-5.03 m/s = (-9.8 m/s2)*tup(-5.03 m/s)/(-9.8 m/s2) = tuptup = 0.513 shang time = 1.03 sReturn to Problem 11 Given:vi = 0 m/svf = 521 m/sd = 0.840 mFind:a = ??vf2 = vi2 + 2*a*d(521 m/s)2 = (0 m/s)2 + 2*(a)*(0.840 m)271441 m2/s2 = (0 m/s)2 + (1.68 m)*a(271441 m2/s2)/(1.68 m) = aa = 1.62*105 m /s2